How many ways are there to arrange the letters of the word GARDEN with the vowels in alphabetical order?
I tried solving it, but I'm not even getting a clue.
So the vowels in "GARDEN" are 'A' and 'E'. Total permutations of the word GARDEN are $6! = 720.$ In half of them, A will occur before E and in the other half of the permutaions, E will occur before A. This is obvious as "A before E" and "E before A" are equally likely and exclusive(nothing other than these two can happen) events.
So both must have probability $1/2$. Hence in half of the words, vowels are in alphabetical order. Hence, answer is $720/2 = 360$.
There are $5+4+3+2+1 = 15$ ways to arrange the vowels $A$, and $E$ in alphabetical order. After that, you have $4! = 24$ ways to arrange the remaining $4$ consonants . Thus the total number of words formed this way is: $15\cdot 24 = 360$
You can start with the combination $AEgrdn$. The consonants can be arranged in $4!=24$ ways. Now the letter $E$ can be put on the positions $2,3,4,5,6$. Thus in total we have $5\cdot 24=120$ with A at position 1 and E behind.
Next $A$ is put on position 2: $gAErdn$. Again the consonants can be arranged in $24$ ways. And $E$ can be put on the positions $3,4,5,6$. Therefore in total we have $4\cdot 24=96$ with $A$ at position $2$ and $E$ behind.
Proceed in this manner till $A$ at position $5$ and $E$ at postion $6$.